Can you solve 8÷2(2+2)

[DoggyDaddy]

I said it's real for 4th grade math

We used the "hockey stick" looking kinda thing for long division in 4th grade.

 

[Jludo]

We used the "hockey stick" looking kinda thing for long division in 4th grade.

Maybe I'm thinking of 2nd grade, it's been a while

 

[eldirector]

8/2(2+2) distributed would be

(8/2*2) + (8/2*2)

You can't cherry pick what parts of the problem you want to "distribute".

Guess what the product is?

 

[ultra...good]

So, when do we get to factoring binomials.
This order of operations stuff is just said.
And for those that really pay attention, what is the oldest number(s)?
Kind if a giveaway, but?

 

[danimal]

8/2(2+2) distributed would be

(8/2*2) + (8/2*2)

You can't cherry pick what parts of the problem you want to "distribute".

Guess what the product is?

I'm not picking what part of the number gets distributed, the division operator does.
The division operator has ownership of 2 parts of a statement, something that exists before it (the numerator), and something that exists after it (the denominator).
Everything before the / is "8"
Everything after it is "2(2+2)"
Is there anything in the before that has rules or properties that supersede the division operator? No, so the numerator is 8.
Is there anything in the after part that has rules or properties that supersede the division operator? Yes, so the denominator requires evaluation before proceeding with the division calculation.
What in the denominator requires evaluation? We have parenthesis and the distributive property. So you must evaluate "2(2+2)" and use it as the denominator.

 

[PistolBob]

If a chicken and a half laid an egg and a half in a day and a half, what would the cost be for 11 2x4's and 3 pounds of doornails?

 

[danimal]

If a chicken and a half laid an egg and a half in a day and a half, what would the cost be for 11 2x4's and 3 pounds of doornails?

At Menards or HomeDepot?

 

[2A_Tom]

$23.57

 

[danimal]

$23.57

Look at Mr. High Brow getting his home improvement supplies from The Sharper Image

 

[Dead Duck]

OK...
I see how it is...

1911
OC
9mm
1

 

[BugI02]

There were no variables in the original problem, so introducing them is a moot point. The original expression contained only integers, so the order of operations still applies.

But I'm using your fave PEDMAS, but the answer appears to be ... wrong. Variables shouldn't matter, we're talking about order of operation

 

[BugI02]

Why would you get xy? It would simplify to x/y. (I can't seem to find a "divided by" symbol on my keyboard other than the slash.)[/QUOTE}

The implied multiplication they're talking about makes that expression 3 * x ÷ 3 * y, and according to them multiplication and division have equivalent priority and so are executed left to right.Thus 3x is divided by 3 to give x which is then multiplied by y to give xy, which is ... not right

 

[qwerty]

100000(this thread) = The number of conversations just like this happening across America.

 

[Dead Duck]

100000(this thread) = The number of conversations just like this happening across America.

Which is exactly why the aliens are laughing at us...... again.



Although....
Mathematics is the universal language so they might be having this same debate.

 

[2A_Tom]

Where are you getting these duck?

 

[Dead Duck]

I steal from other forums and post like a picture.

 

[Snapdragon]

But I'm using your fave PEDMAS, but the answer appears to be ... wrong. Variables shouldn't matter, we're talking about order of operation

But once you introduce variable terms, the coefficient and the variable become one unit that you cannot separate willy-nilly. 6/2x is not (six divided by two) times x, or 3x, because you cannot separate the 2 from the x and only divide by the 2 (unless it specifically says (6/2)x. 6/2x is 6 divided by the 2x term, which we can simplfy to 3/x. Whole different ball game. Order of operations is still followed, but variable terms hold together as one unit unless there is a multiplication operator between the number and the variable. In this case, the number ceases to act as the coefficient because in the expression 6/2*x, the coefficient of x is 1.

 

[2A_Tom]

There are other forums?

 

[danimal]

The implied multiplication they're talking about makes that expression 3 * x ÷ 3 * y, and according to them multiplication and division have equivalent priority and so are executed left to right. Thus 3x is divided by 3 to give x which is then multiplied by y to give xy, which is ... not right

I was always of the impression that a number or variable proceeding the parenthesis belonged to the parenthesis as though there was an implied outer pair of parenthesis like 8/[2(2+2)] because of the distributive property. Not unlike your example of numbers that precede a variable; there is an implied () around them because they are treated as a single element each of the division operator (num and dom). And until some other operation comes along to eliminate or simplify the distributor or distributee, they stay together.

I personally don't see a multiplication operator between the 2 and the grouping, when I read it I see distribute the 2. The reason why I don't see 8/2 as the distributor is because of the division operator. If I have a portion of a statement that reads "8y/2", you simplify to 4y right? What if it's "8/2y", do you still simplify to 4y? Is "8/x^2" equal to "8 divided by x then squared", or "8 divided by the square of x"? If you have a fraction preceding something like a variable, parenthesis, or exponent, I always treat it as part of the denominator, essentially that's how I'm reading it.

But once you introduce variable terms, the coefficient and the variable become one unit that you cannot separate willy-nilly. 6/2x is not (six divided by two) times x, or 3x, because you cannot separate the 2 from the x and only divide by the 2 (unless it specifically says (6/2)x. 6/2x is 6 divided by the 2x term, which we can simplfy to 3/x. Whole different ball game. Order of operations is still followed, but variable terms hold together as one unit unless there is a multiplication operator between the number and the variable. In this case, the number ceases to act as the coefficient because in the expression 6/2*x, the coefficient of x is 1.

I see it going either way lacking a real equation to give more context. As presented I may be reading too much into it or processing it too thoroughly with treating the grouping like a variable substitution as part of the denominator. I'm just used to it from working out my own mathematical equations where I work. You always know how to solve your own equations... once you figure them out. Homeschooling my boys we haven't encountered anything this ambiguous in my oldest son's first year of algebra. There's usually more or the question is crafted to be more specific.

There are other forums?

Lies, don't listen to him

 

[BugI02]

But once you introduce variable terms, the coefficient and the variable become one unit that you cannot separate willy-nilly. 6/2x is not (six divided by two) times x, or 3x, because you cannot separate the 2 from the x and only divide by the 2 (unless it specifically says (6/2)x. 6/2x is 6 divided by the 2x term, which we can simplfy to 3/x. Whole different ball game. Order of operations is still followed, but variable terms hold together as one unit unless there is a multiplication operator between the number and the variable. In this case, the number ceases to act as the coefficient because in the expression 6/2*x, the coefficient of x is 1.

So why isn't 2 the coefficient of 2(2+2), shouldn't the general form give the same answer as the specific if PEDMAS is the only thing that needs to be taken into consideration? 2(2+2) is one term that needs to be completely simplified before proceeding to divide

Jetta is the rightest poster, it is sloppy to write the expression with ambiguity and the author needs a dope slap